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Photoelectric emission

  1. Nov 10, 2009 #1
    It is known that for light of a given intensity when incident on a photoelectric material, the number of electrons emitted is independent of the frequency of light, but increases when the intensity of light goes up. However, the maximum kinetic energy of the emitted electrons increases with the frequency of incident light {KE (max) = hNu - W}. [Nu = frequency of light, and W = work function of the photoelectric material]. Therefore the velocity of the electrons increases with frequency of incident light.
    I would like to understand as to why the number of electrons emitted is independent of frequency even though the maximum kinetic energy, and hence velocity of emitted electrons, goes up with frequency. I shall appreciate help on this score. Thanks.
     
  2. jcsd
  3. Nov 10, 2009 #2

    mathman

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    Einstein won his Nobel prize for answering your question. The essential point is that light is considered to be made up of discrete entities (photons). One photon, assuming above the minimum energy, can knock out only one electron, no matter how much energy the photon has.
     
  4. Nov 10, 2009 #3
    Thank you for your post. However please consider the following:

    Increasing the frequency increases the energy of the photons. Therefore, if the frequency is increased while keeping the intensity of radiation (the total photon energy per unit area per unit time) the same, the number of photons striking the photoelectric material per unit area per unit time will decrease because the energy of each photon goes up. Hence the number of photoelectrons emitted by the photoelectric material should also decrease with increase in frequency of the incident light without changing the intensity. I am, therefore, not sure as to whether the photoelectric current (number of emitted photoelectrons per unit time) is really independent of the frequency of the incident radiation when the intensity of light remains constant. Maybe you could look at the issue from this angle. Thanks.
     
  5. Nov 11, 2009 #4
    Errr..no. When the frequency (and the energy) of the light is too low to emit electrons with sufficient energy to exceed the work function, no electrons are emitted.
    Bob S
     
  6. Nov 11, 2009 #5

    mathman

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    If you increase the energy per photon (increase frequency) while keeping the total energy constant, thus decreasing the photon rate, you will decrease the electron emission rate, but increase the energy per electron. Why not??
     
  7. Nov 11, 2009 #6
    You are correct, the number of photoelectrons per unit energy (eV) decreases, because the photoelectric effect is a point interaction between one photon and one electron. The excess energy just goes into electron kinetic energy. For blue light on a solar cell, there is enough excess energy to produce two photo-electrons rather than just one. Solar cell developers are working to get two electrons for short wavelength light. Note: "Photoelectrons" in solar cells are electrons kicked up to the conduction band, not "free" electrons ejected from the material.
    Bob S
     
    Last edited: Nov 11, 2009
  8. Nov 11, 2009 #7
    To summarise:

    (a) For a given intensity of light, the rate of emission of photoelectrons, and hence the photoelectric current, decreases with increasing frequency of the irradiating source, provided th energy per photon exceeds the work function of the photoelectric substance.

    (b) However, the kinetic energy of the photoelectrons, and hence the energy per electron, goes up with increasing frequency of the incident light for frequencies that exceed the threshold frequency.
     
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